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How can you find the value of a square root number? - Answers

First, use the identity: sqrt(x)=e^[ln(x)/2]Then, calculate ln(x) by using the series:2*SUM({[1/(2n+1)]*[(x-1)/(x+1)]}^(2n+1),n,0,infinity)We'll call the above number yThen calculate e^(y/2) via the expansion e^(y/2)=SUM{[(y/2)^n]/n!,n,0,infinity}Since it's actually impossible to expand a number out to infinity by hand, you can almost never get the exact value of sqrt(x), but the larger the upper boundary of n you use, the more accurate the approximation.



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How can you find the value of a square root number? - Answers

https://math.answers.com/math-and-arithmetic/How_can_you_find_the_value_of_a_square_root_number

First, use the identity: sqrt(x)=e^[ln(x)/2]Then, calculate ln(x) by using the series:2*SUM({[1/(2n+1)]*[(x-1)/(x+1)]}^(2n+1),n,0,infinity)We'll call the above number yThen calculate e^(y/2) via the expansion e^(y/2)=SUM{[(y/2)^n]/n!,n,0,infinity}Since it's actually impossible to expand a number out to infinity by hand, you can almost never get the exact value of sqrt(x), but the larger the upper boundary of n you use, the more accurate the approximation.



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https://math.answers.com/math-and-arithmetic/How_can_you_find_the_value_of_a_square_root_number

How can you find the value of a square root number? - Answers

First, use the identity: sqrt(x)=e^[ln(x)/2]Then, calculate ln(x) by using the series:2*SUM({[1/(2n+1)]*[(x-1)/(x+1)]}^(2n+1),n,0,infinity)We'll call the above number yThen calculate e^(y/2) via the expansion e^(y/2)=SUM{[(y/2)^n]/n!,n,0,infinity}Since it's actually impossible to expand a number out to infinity by hand, you can almost never get the exact value of sqrt(x), but the larger the upper boundary of n you use, the more accurate the approximation.

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      First, use the identity: sqrt(x)=e^[ln(x)/2]Then, calculate ln(x) by using the series:2*SUM({[1/(2n+1)]*[(x-1)/(x+1)]}^(2n+1),n,0,infinity)We'll call the above number yThen calculate e^(y/2) via the expansion e^(y/2)=SUM{[(y/2)^n]/n!,n,0,infinity}Since it's actually impossible to expand a number out to infinity by hand, you can almost never get the exact value of sqrt(x), but the larger the upper boundary of n you use, the more accurate the approximation.

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